Adjacency List Representation in DSA C - Time & Space Complexity

Time Complexity: Adjacency List Representation

O(n + e)

Understanding Time Complexity

We want to understand how the time to build and use an adjacency list changes as the graph grows.

How does the number of nodes and edges affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Create adjacency list for graph with n vertices and edges array
void createAdjList(int n, int edges[][2], int edgesCount, List adjList[]) {
    for (int i = 0; i < n; i++) {
        adjList[i] = createEmptyList();
    }
    for (int i = 0; i < edgesCount; i++) {
        int u = edges[i][0];
        int v = edges[i][1];
        addToList(adjList[u], v);
        addToList(adjList[v], u); // for undirected graph
    }
}
    

This code builds an adjacency list for an undirected graph by adding each edge to the lists of both connected nodes.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Loop over all edges to add connections.
• How many times: The loop runs once for each edge, twice adding nodes to lists.

How Execution Grows With Input

As the number of edges grows, the work grows roughly twice as fast because each edge adds two entries.

Input Size (edgesCount)	Approx. Operations
10	About 20 additions to lists
100	About 200 additions to lists
1000	About 2000 additions to lists

Pattern observation: The operations grow linearly with the number of edges.

Final Time Complexity

Time Complexity: O(n + e)

This means the time grows linearly with the number of nodes plus the number of edges.

Common Mistake

[X] Wrong: "The time depends only on the number of nodes (n)."

[OK] Correct: The edges (connections) also matter because each edge adds work to the lists.

Interview Connect

Understanding adjacency list complexity helps you explain graph storage and traversal efficiency clearly in interviews.

Self-Check

"What if the graph was directed and we only added one direction per edge? How would the time complexity change?"

Practice

1. What does an adjacency list represent in graph data structures?

easy

A. A list where each vertex stores its adjacent vertices

B. A matrix showing connections between vertices

C. A list of all vertices without edges

D. A list of edges only without vertices

Solution

1. Step 1: Understand adjacency list concept

   An adjacency list stores each vertex and the vertices it connects to directly.

2. Step 2: Compare with other representations

   Unlike adjacency matrix, adjacency list focuses on neighbors per vertex, not a full matrix.

3. Final Answer:

   A list where each vertex stores its adjacent vertices -> Option A

4. Quick Check:

   Adjacency list = vertex with neighbors [OK]

Hint: Adjacency list stores neighbors per vertex, not full matrix [OK]

Common Mistakes:

• Confusing adjacency list with adjacency matrix
• Thinking it stores only edges without vertices
• Assuming it lists vertices without connections

2. Which of the following is the correct C syntax to declare an adjacency list for a graph with 5 vertices using an array of pointers to linked lists?

easy

A. int* adjList[5];

B. int adjList[5][5];

C. struct Node adjList[5];

D. struct Node* adjList[5];

Solution

1. Step 1: Understand adjacency list declaration

   Adjacency list uses array of pointers to linked list nodes, so 'struct Node* adjList[5];' is correct.

2. Step 2: Check other options

   int adjList[5][5]; is adjacency matrix, C is array of structs (not pointers), A is array of int pointers (not nodes).

3. Final Answer:

   struct Node* adjList[5]; -> Option D

4. Quick Check:

   Array of pointers to nodes = correct adjacency list declaration [OK]

Hint: Adjacency list uses array of pointers to node structs [OK]

Common Mistakes:

• Using 2D array instead of array of pointers
• Declaring array of structs instead of pointers
• Using int pointers instead of node pointers

3. Given the following adjacency list representation code snippet for a graph with 3 vertices (0,1,2):

struct Node {
  int vertex;
  struct Node* next;
};

struct Node* adjList[3] = {NULL, NULL, NULL};

// Adding edge 0->1
struct Node* newNode = malloc(sizeof(struct Node));
newNode->vertex = 1;
newNode->next = adjList[0];
adjList[0] = newNode;

// Adding edge 0->2
newNode = malloc(sizeof(struct Node));
newNode->vertex = 2;
newNode->next = adjList[0];
adjList[0] = newNode;

// Adding edge 1->2
newNode = malloc(sizeof(struct Node));
newNode->vertex = 2;
newNode->next = adjList[1];
adjList[1] = newNode;

// Print adjacency list for vertex 0
struct Node* temp = adjList[0];
while(temp != NULL) {
  printf("%d -> ", temp->vertex);
  temp = temp->next;
}
printf("NULL\n");

What is the printed output?

medium

A. 1 -> 2 -> NULL

B. 0 -> 1 -> 2 -> NULL

C. 2 -> 1 -> NULL

D. NULL

Solution

1. Step 1: Trace insertion order for vertex 0

   Edges 0->1 then 0->2 are added at head, so list is 2 -> 1 -> NULL.

2. Step 2: Understand print loop

   Prints vertices in order from head: 2 then 1, then NULL.

3. Final Answer:

   2 -> 1 -> NULL -> Option C

4. Quick Check:

   Last inserted edge appears first in adjacency list [OK]

Hint: New nodes added at head, so last added appears first [OK]

Common Mistakes:

• Assuming order is insertion order instead of reversed
• Printing vertex 0 itself in adjacency list
• Confusing NULL print format

4. Consider this buggy code snippet to add an edge from vertex u to v in an adjacency list:

void addEdge(struct Node* adjList[], int u, int v) {
  struct Node* newNode = malloc(sizeof(struct Node));
  newNode->vertex = v;
  newNode->next = NULL;
  adjList[u] = newNode;
}

What is the main problem with this code?

medium

A. It overwrites existing adjacency list for vertex u, losing previous edges

B. It does not allocate memory for newNode

C. It sets newNode->next incorrectly to itself

D. It does not assign vertex v to newNode

Solution

1. Step 1: Analyze pointer assignment

   Setting adjList[u] = newNode overwrites old list head, losing previous edges.

2. Step 2: Check linked list linking

   newNode->next is NULL, so no link to old list; old edges lost.

3. Final Answer:

   It overwrites existing adjacency list for vertex u, losing previous edges -> Option A

4. Quick Check:

   Overwriting head pointer loses old adjacency list [OK]

Hint: Always link new node to old head to keep edges [OK]

Common Mistakes:

• Not linking new node to existing list
• Forgetting to allocate memory
• Incorrectly setting next pointer

5. You have a directed graph with 4 vertices (0 to 3). You want to represent it using adjacency lists and add edges:
- 0 -> 1
- 0 -> 2
- 1 -> 2
- 2 -> 0
- 2 -> 3
- 3 -> 3

Which data structure and insertion method correctly represents this graph in C?

hard

A. Use 2D int array; set 1 for edges and 0 otherwise

B. Use array of pointers to Node structs; insert new nodes at head linking to old list

C. Use array of Node structs; overwrite each vertex's list on insertion

D. Use array of int pointers; store vertex numbers without linking

Solution

1. Step 1: Choose correct data structure

   Adjacency list uses array of pointers to linked nodes for each vertex.

2. Step 2: Insert edges properly

   Insert new nodes at head, linking to previous list to keep all edges.

3. Step 3: Check other options

   Use 2D int array; set 1 for edges and 0 otherwise is adjacency matrix, C overwrites lists losing edges, D lacks linking.

4. Final Answer:

   Use array of pointers to Node structs; insert new nodes at head linking to old list -> Option B

5. Quick Check:

   Array of pointers + head insertion = correct adjacency list [OK]

Hint: Use pointers array and link new nodes at head to keep edges [OK]

Common Mistakes:

• Using adjacency matrix instead of list
• Overwriting adjacency list on insertion
• Not linking nodes properly